Question

Consider the following compounds:

KO₂, H₂O₂, and H₂SO₄

The oxidation states of the underlined elements in them are, respectively:

Options

• +1, −1, and +6

• +2, −2, and +6

• +1, −2, and +4

• +4, −4, and +6

Answer 1

+1, −1, and +6

Explanation:

Oxidation state of K in KO₂

K is an alkali metal that always shows a +1 oxidation state in compounds.

Oxidation state of O in H₂O₂

2(+1) + 2x = 0 

2 + 2x = 0

2x = −2

x = −1 

Oxidation state of S in H₂SO₄

2(+1) + x + 4(−2) = 0

2 + x + (−8) = 0

x = 8 − 2

x = +6